Problem: Find the slope and y-intercept of the line that is ${\text{parallel}}$ to $\enspace {y = \dfrac{3}{2}x + 2}\enspace$ and passes through the point ${(-7, -5)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Parallel lines have the same slope. The slope of the blue line is ${\dfrac{3}{2}}$ , so the equation of our parallel line will be of the form $\enspace {y = \dfrac{3}{2}x + b}\enspace$ We can plug our point, $(-7, -5)$ , into this equation to solve for ${b}$ , the y-intercept. $-5 = {\dfrac{3}{2}}(-7) + {b}$ $-5 = -\dfrac{21}{2} + {b}$ $-5 + \dfrac{21}{2} = {b} = \dfrac{11}{2}$ The equation of the parallel line is $\enspace {y = \dfrac{3}{2}x + \dfrac{11}{2}}\enspace$. ${m = \dfrac{3}{2}, \enspace b = \dfrac{11}{2}}$